Largest product of a subarray of size k

Given an array consisting of n positive integers, and an integer k. Find the largest product subarray of size k, i.e., find maximum produce of k contiguous elements in the array where k <= n.

Examples :

Input: arr[] = {1, 5, 9, 8, 2, 4,
                 1, 8, 1, 2} 
       k = 6
Output:   4608  
The subarray is {9, 8, 2, 4, 1, 8}

Input: arr[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}
       k = 4
Output:   720  
The subarray is {5, 9, 8, 2}

Input: arr[] = {2, 5, 8, 1, 1, 3};
       k = 3             
Output:   80  
The subarray is {2, 5, 8}

Method 1 (Simple : O(n*k))
A Naive approach would be to consider all the subarrays of size k one by one. Such a approach would require two loops hence the complexity would be O(n*k).

Method 2 (Efficient : O(n))
We can solve it in O(n) by using the fact that product of a subarray of size k can be computed in O(1) time if we have product of previous subarray available with us.



curr_product = (prev_product / arr[i-1]) * arr[i + k -1]

prev_product : Product of subarray of size k beginning 
               with arr[i-1]

curr_product : Product of subarray of size k beginning 
               with arr[i]

In this way we can compute the maximum k size subarray product in only one traversal. Below is C++ implementation of the idea.

C++

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// C++ program to find the maximum product of a subarray
// of size k.
#include <bits/stdc++.h>
using namespace std;
  
// This function returns maximum product of a subarray
// of size k in given arrar, arr[0..n-1]. This function
// assumes that k is smaller than or equal to n.
int findMaxProduct(int arr[], int n, int k)
{
    // Initialize the MaxProduct to 1, as all elements
    // in the array are positive
    int MaxProduct = 1;
    for (int i=0; i<k; i++)
        MaxProduct *= arr[i];
  
    int prev_product = MaxProduct;
  
    // Consider every product beginning with arr[i]
    // where i varies from 1 to n-k-1
    for (int i=1; i<=n-k; i++)
    {
        int curr_product = (prev_product/arr[i-1]) *
                            arr[i+k-1];
        MaxProduct = max(MaxProduct, curr_product);
        prev_product = curr_product;
    }
  
    // Return the maximum product found
    return MaxProduct;
}
  
// Driver code
int main()
{
    int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2};
    int k = 6;
    int n = sizeof(arr1)/sizeof(arr1[0]);
    cout << findMaxProduct(arr1, n, k) << endl;
  
    k = 4;
    cout << findMaxProduct(arr1, n, k) << endl;
  
    int arr2[] = {2, 5, 8, 1, 1, 3};
    k = 3;
    n = sizeof(arr2)/sizeof(arr2[0]);
    cout << findMaxProduct(arr2, n, k);
  
    return 0;
}

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Java














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// Java program to find the maximum product of a subarray 


// of size k 


import java.io.*; 


import java.util.*; 


  


class GFG  


{ 


    // Function returns maximum product of a subarray 


    // of size k in given arrar, arr[0..n-1]. This function 


    // assumes that k is smaller than or equal to n. 


    static int findMaxProduct(int arr[], int n, int k) 


    { 


        // Initialize the MaxProduct to 1, as all elements 


        // in the array are positive 


        int MaxProduct = 1; 


        for (int i=0; i<k; i++) 


            MaxProduct *= arr[i]; 


   


        int prev_product = MaxProduct; 


   


        // Consider every product beginning with arr[i] 


        // where i varies from 1 to n-k-1 


        for (int i=1; i<=n-k; i++) 


        { 


            int curr_product = (prev_product/arr[i-1]) * 


                                arr[i+k-1]; 


            MaxProduct = Math.max(MaxProduct, curr_product); 


            prev_product = curr_product; 


        } 


   


        // Return the maximum product found 


        return MaxProduct; 


    } 


      


    // driver program 


    public static void main (String[] args)  


    { 


        int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}; 


        int k = 6; 


        int n = arr1.length; 


        System.out.println(findMaxProduct(arr1, n, k)); 


   


        k = 4; 


        System.out.println(findMaxProduct(arr1, n, k)); 


   


        int arr2[] = {2, 5, 8, 1, 1, 3}; 


        k = 3; 


        n = arr2.length; 


        System.out.println(findMaxProduct(arr2, n, k)); 


    } 


} 


  


// This code is contributed by Pramod Kumar 



















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Python3














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# Python 3 program to find the maximum  


# product of a subarray of size k. 


  


# This function returns maximum product  


# of a subarray of size k in given arrar, 


# arr[0..n-1]. This function assumes  


# that k is smaller than or equal to n. 


def findMaxProduct(arr, n, k) : 


    


    # Initialize the MaxProduct to 1,  


    # as all elements in the array  


    # are positive 


    MaxProduct = 1


    for i in range(0, k) : 


        MaxProduct = MaxProduct * arr[i] 


          


    prev_product = MaxProduct 


   


    # Consider every product beginning 


    # with arr[i] where i varies from 


    # 1 to n-k-1 


    for i in range(1, n - k + 1) : 


        curr_product = (prev_product // arr[i-1]) * arr[i+k-1] 


        MaxProduct = max(MaxProduct, curr_product) 


        prev_product = curr_product 


      


      


    # Return the maximum product found 


    return MaxProduct 


      


# Driver code 


arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2] 


k = 6


n = len(arr1) 


print (findMaxProduct(arr1, n, k) ) 


k = 4


print (findMaxProduct(arr1, n, k)) 


  


arr2 = [2, 5, 8, 1, 1, 3] 


k = 3


n = len(arr2) 


  


print(findMaxProduct(arr2, n, k)) 


  


# This code is contributed by Nikita Tiwari. 



















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C#














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// C# program to find the maximum  


// product of a subarray of size k 


using System; 


  


class GFG  


{ 


    // Function returns maximum  


    // product of a subarray of  


    // size k in given arrar,  


    // arr[0..n-1]. This function  


    // assumes that k is smaller  


    // than or equal to n. 


    static int findMaxProduct(int []arr,  


                              int n, int k) 


    { 


        // Initialize the MaxProduct  


        // to 1, as all elements 


        // in the array are positive 


        int MaxProduct = 1; 


        for (int i = 0; i < k; i++) 


            MaxProduct *= arr[i]; 


  


        int prev_product = MaxProduct; 


  


        // Consider every product beginning  


        // with arr[i] where i varies from  


        // 1 to n-k-1 


        for (int i = 1; i <= n - k; i++) 


        { 


            int curr_product = (prev_product /  


                                 arr[i - 1]) *  


                                 arr[i + k - 1]; 


            MaxProduct = Math.Max(MaxProduct,  


                                  curr_product); 


            prev_product = curr_product; 


        } 


  


        // Return the maximum 


        // product found 


        return MaxProduct; 


    } 


      


    // Driver Code 


    public static void Main ()  


    { 


        int []arr1 = {1, 5, 9, 8, 2,  


                      4, 1, 8, 1, 2}; 


        int k = 6; 


        int n = arr1.Length; 


        Console.WriteLine(findMaxProduct(arr1, n, k)); 


  


        k = 4; 


        Console.WriteLine(findMaxProduct(arr1, n, k)); 


  


        int []arr2 = {2, 5, 8, 1, 1, 3}; 


        k = 3; 


        n = arr2.Length; 


        Console.WriteLine(findMaxProduct(arr2, n, k)); 


    } 


} 


  


// This code is contributed by anuj_67. 



















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PHP














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<?php 


// PHP program to find the maximum  


// product of a subarray of size k. 


  


// This function returns maximum  


// product of a subarray of size  


// k in given arrar, arr[0..n-1]. 


// This function assumes that k  


// is smaller than or equal to n. 


function findMaxProduct( $arr, $n, $k) 


{ 


      


    // Initialize the MaxProduct to 


    // 1, as all elements 


    // in the array are positive 


    $MaxProduct = 1; 


    for($i = 0; $i < $k; $i++) 


        $MaxProduct *= $arr[$i]; 


  


    $prev_product = $MaxProduct; 


  


    // Consider every product 


    // beginning with arr[i] 


    // where i varies from 1  


    // to n-k-1 


    for($i = 1; $i < $n - $k; $i++) 


    { 


        $curr_product = ($prev_product / $arr[$i - 1]) * 


                                       $arr[$i + $k - 1]; 


        $MaxProduct = max($MaxProduct, $curr_product); 


        $prev_product = $curr_product; 


    } 


  


    // Return the maximum 


    // product found 


    return $MaxProduct; 


} 


  


    // Driver code 


    $arr1 = array(1, 5, 9, 8, 2, 4, 1, 8, 1, 2); 


    $k = 6; 


    $n = count($arr1); 


    echo findMaxProduct($arr1, $n, $k),"\n" ; 


  


    $k = 4; 


    echo findMaxProduct($arr1, $n, $k),"\n"; 


  


    $arr2 = array(2, 5, 8, 1, 1, 3); 


    $k = 3; 


    $n = count($arr2); 


    echo findMaxProduct($arr2, $n, $k); 


  


// This code is contributed by anuj_67. 


?> 



















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Output :


4608
720
80



This article is contributed by Ashutosh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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